Solve (1/3)b^(3)+2b=15
I don't know how to do/_\
Is it use factor formula? Or some methods taught in s.1-s.3?
Please help, thanks!
Maths S.1-S.3 question
2014-12-01 12:31 am
回答 (3)
2014-12-01 4:08 am
✔ 最佳答案
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2014-12-01 4:18 am
岩岩睇錯仲以為係 ² ~.~
2014-12-01 12:34 am
I think so, factor theorem, or trial-and-error.
The answer is b = 3.
http://www.wolframalpha.com/input/?i=(1%2F3)b%5E(3)%2B2b%3D15&t=crmtb01
2014-11-30 16:37:02 補充：
b³/3 + 2b = 15
b³ + 6b = 45
b³ + 6b - 45 = 0
Let f(b) = b³ + 6b - 45 is a polynomial in b.
Try f(3) = 3³ + 6(3) - 45 = 0
(b - 3) is a factor of f(b).
b³ + 6b - 45 = 0
(b - 3)(b² + 3b + 15) = 0
The answer is b = 3.
http://www.wolframalpha.com/input/?i=(1%2F3)b%5E(3)%2B2b%3D15&t=crmtb01
2014-11-30 16:37:02 補充：
b³/3 + 2b = 15
b³ + 6b = 45
b³ + 6b - 45 = 0
Let f(b) = b³ + 6b - 45 is a polynomial in b.
Try f(3) = 3³ + 6(3) - 45 = 0
(b - 3) is a factor of f(b).
b³ + 6b - 45 = 0
(b - 3)(b² + 3b + 15) = 0
收錄日期: 2021-04-15 17:25:34
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